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vredina [299]
3 years ago
9

Choose the correct statement regarding projectile motion in the absence of air resistance. Assume the object is at sea level mov

ing from left to right.
The magnitude of the acceleration in the x direction is always zero.
At the apex, in the y direction, the velocity is zero and the acceleration is positive.
At the apex, in the x direction, the velocity is zero and the acceleration is zero.
The magnitude of the velocity in the y direction is always constant.
At the apex, in the y direction, the velocity is negative and the acceleration is zero.
Physics
1 answer:
pishuonlain [190]3 years ago
7 0

Answer:

Explanation:

The magnitude of the acceleration in the x direction is always zero: TRUE.

At the apex, in the y direction the velocity is zero and the accelration is positive. TRUE.

At the apex, in the x direction, the velocity is zero and the acceleration is zero. FALSE. The accelration is zero, but the velocity is the same it had when it was shot.

The magnitude of the velocity in the y direction is always constant. FALSE, it's subject to gravity and it's velocity varies as v=v_0 - 9.81ms^{-2} t

At the apex, in the Y direction, the velocity is negative and the acceleration is zero. FALSE. Velocity is zero, Acceleration is 9.81 ms^{-2}, towards the negative y axis

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I need to find the current resistance and voltage for each in this complicated circuit plz help
konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel.  That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

7 0
3 years ago
(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the per
Blizzard [7]

Answer:

\huge\boxed{T = 0.025\ seconds}

Explanation:

<u>Given:</u>

Frequency = f = 40 Hz

<u>Required:</u>

Time period = T = ?

<u>Formula:</u>

\sf T = 1 / f

<u>Solution:</u>

T = 1 / 40

T = 0.025 seconds

\rule[225]{225}{2}

Hope this helped!

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Two golfers each hit a ball at the same speed, but one at 60 degrees with the horizontal and the other 30 degrees. a. Which ball
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Answer:

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

Explanation:

Both balls will hit the ground at the same distance and at the same time assuming they both start at the same initial elevation from the ground.

We know that range in the projectile motion given as

R=\dfrac{v_o^2sin2\theta}{g}

sin 60° = sin 120° =0.866

Therefore, the maximum range will be identical assuming both golfers hit the ball at the same elevation from the ground.

As for the time...

Where v is 0 m/s (final velocity at highest point), v0 is the initial velocity (for both golfers) and is the acceleration due to gravity (-9.8 m/s^²). As you can see, the time it takes to get to the highest point is independent of the angle; it is only dependent on the initial velocity. Since both golfers hit the ball with the same speed, the time for the ball to reach the highest point will be the same. Also, since it's a parabola, you can multiply the time by 2 for each golfer to get the time it takes to hit the ground.

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

5 0
3 years ago
An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust
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Answer:

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Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

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